Re: Subroutine foo redefined a bar

On Tue, Jul 28, 2009 at 17:05, Dermot<paikkos@xxxxxxxxxxxxxx> wrote:
snip
>> Â Âmy $bid = shift;
>> Â Âmy $items = ref $_[0] ? $_[0] : \@_;
>>
>
> Perhaps you can expand, if $_[0] was a scalar wouldn't that get
> assigned to $items?
snip

Normal scalars (i.e. strings and numbers) are not references,
therefore the result of the ref call will be undef and \@_ will be
used.  If $_[0] is a reference it will be used.

snip
> I venture this but only because I suspect I am missing something in
> your example:
> my $items = ( ref($_[0]) eq 'ARRAY' ) ? $_[0] : \@_;
snip

Never say ref($foo) eq 'ARRAY', it is extraordinarily dangerous.  If
$foo is a blessed array ref you will get back the class it has been
blessed into rather than 'ARRAY'.  It is safe to use ref to find out
two things: if a scalar is a reference and the class of an object.
Use the reftype function from [Scalar::Util][1] if you want to know
what type of variable a reference points to.

For maximum safety, you should write the code like this

use Carp;
use Scalar::Util qw/reftype/;

.
.
.

sub addItemsToBasket {
    croak "bad number of arguments" unless @_ == 2;
    my ($bid, $items) = @_;

    #code to check that $bid is a valid value, croak if it isn't

    $items = [$items] unless ref $items;
    croak "ITEMS should be an arrayref or a simple scalar"
        unless reftype $items eq "ARRAY";

    for my $item (@$items) {
         unless (SOME CHECK TO MAKE SURE ITEM IS VALID) {
             carp "item $item is not valid";
             next;
         }

         #do stuff with $item
    }
}


[1] : http://perldoc.perl.org/Scalar/Util.html



-- 
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.

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